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stich3 [128]
3 years ago
8

-6 + x over 4=-5 what is the value of x

Mathematics
2 answers:
My name is Ann [436]3 years ago
6 0

Answer:x=26

Step-by-step explanation:

olga nikolaevna [1]3 years ago
5 0

Answer:

I am kinda good at this I think the answer is 26 correct me if I am wrong

Step-by-step

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WARRIOR [948]

(x+3)(x-5)=x^2-5x+3x-15=x^2-2x-15

6 0
4 years ago
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Please help! Thanks!
Over [174]
It's c u do distribute the -4 to everything in the ( )'s then combine the like terms
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3 years ago
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A player of a video game is confronted with a series of four opponents and an 80% probability of defeating each opponent. Assume
mixer [17]

Answer:

(a) 0.4096

(b) 0.64

(c) 0.7942

Step-by-step explanation:

The probability that the player wins is,

P(W)=0.80

Then the probability that the player losses is,

P(L)=1-P(W)=1-0.80=0.20

The player is playing the video game with 4 different opponents.

It is provided that when the player is defeated by an opponent the game ends.

All the possible ways the player can win is: {L, WL, WWL, WWWL and WWWW)

(a)

The results from all the 4 opponents are independent, i.e. the result of a game played with one opponent is unaffected by the result of the game played with another opponent.

The probability that the player defeats all four opponents in a game is,

P (Player defeats all 4 opponents) = P(W)\times P(W)\times P(W)\times P(W)=[P(W)]^{4} =(0.80)^{4}=0.4096

Thus, the probability that the player defeats all four opponents in a game is 0.4096.

(b)

The probability that the player defeats at least two opponents in a game is,

P (Player defeats at least 2) = 1 - P (Player losses the 1st game) - P (Player losses the 2nd game) = 1-P(L)-P(WL)

                                    =1-(0.20)-(0.80\times0.20)\\=1-0.20-0.16\\=0.64

Thus, the probability that the player defeats at least two opponents in a game is 0.64.

(c)

Let <em>X</em> = number of times the player defeats all 4 opponents.

The probability that the player defeats all four opponents in a game is,

P(WWWW) = 0.4096.

Then the random variable X\sim Bin(n=3, p=0.4096)

The probability distribution of binomial is:

P(X=x)={n\choose x}p^{x} (1-p)^{n-x}

The probability that the player defeats all the 4 opponents at least once is,

P (<em>X</em> ≥ 1) = 1 - P (<em>X</em> < 1)

             = 1 - P (<em>X</em> = 0)

             =1-[{3\choose 0}(0.4096)^{0} (1-0.4096)^{3-0}]\\=1-[1\times1\times (0.5904)^{3}\\=1-0.2058\\=0.7942

Thus, the probability that the player defeats all the 4 opponents at least once is 0.7942.

3 0
3 years ago
2x-1=10 what is the answer please someone help
Jlenok [28]
10+1=11
2x=11
x=11/2
x=5.5
8 0
3 years ago
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Is there a square made up of square tiles that has only 75 squares overall? What about one that has 120 squares overall? If it i
Mama L [17]

Answer:

See below.

Step-by-step explanation:

There are two essential pieces of information missing in the question, namely:

(1) "can the the square tiles be of varying size?"

(2) "what are the number constraints on the size of a tile?" (integers only? fractional lengths? Any real number?)

Without knowing the above, I can only speculate:

If the answer is "no, the square tiles are all same size, and must be of integer size" then the answer to your question will be "no, neither with 75 nor with 120 tiles a square of integer size can be covered" This is because the numbers 75 and 120 are not perfect squares (unlike, say, 121), which is easy to check.

If the answer is "yes, the square tiles can vary in size, but must be integers" then the answer to your question will be "yes" at least in the case of the 75 tiles...I could come up with an arrangement to solve that and am positive there is one for 120. I am not going into the detail here not knowing what is really required.

If the answer is "tile size can be any real number" then the answer to your question will be "yes" (albeit, there may never be possible to construct).

Let me know if you have questions.

7 0
3 years ago
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